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May 24 edit

Integral of different type edit

Today, I saw a different type of integral sign in electromagnetism. Actually it was a simple integral sign, but it had an additional small circle at the center of the integral sign. I don't know its name, where and how to use it. But I know how to use the simple integral. Someone, please, tell me how to integrate a function using the different type of integral and also give a list of formulas of this integral. Thank you! Concepts of Physics (talk) 02:40, 24 May 2013 (UTC)[reply]

This means that you are taking the (line) integral over a closed loop, or less-commonly (in which case I prefer a double integral sign, but a single one is often used), a closed surface (as in Maxwell's first equation). This is vector calculus, which is best understood with knowledge of multivariable calculus.--Jasper Deng (talk) 02:49, 24 May 2013 (UTC)[reply]

See also contour integral, and e.g. Green's theorem, divergence theorem,... --CiaPan (talk) 05:08, 24 May 2013 (UTC)[reply]

Does this contour integral follow the same rules of integration as followed by simple integral or it follow some different kind of rules? Concepts of Physics (talk) 02:40, 25 May 2013 (UTC)[reply]

If you saw it in electromagnetism, it would almost certainly be a line, surface, or volume integral. The fundamental theorem of calculus does not immediately apply to these, but the theorems linked by CiaPan above assist greatly with evaluation. Of particular usefulness to you would be to have an idea of how to evaluate multiple integrals, as that is how the evaluation theorems for these integrals work. Of course, before you do this, be sure to have a firm grasp of single-variable calculus, especially with regards to the definition of an integral as being the limit of a Riemann sum.

This really is only needed if you must make calculations. The idea of any integral is to subdivide whatever region you are taking the integral over into infinitesimally small pieces, evaluate a particular function for each piece, and add it all up (strictly speaking, we take the limit as the piece size approaches zero).--Jasper Deng (talk) 06:41, 25 May 2013 (UTC)[reply]

Antiderivative of gamma function edit

What is $\int \Gamma (x)\operatorname {d} \!x$ ? $\int _{1}^{x}\Gamma (t)\operatorname {d} \!t$ would also answer my question. Can we take advantage of the fact that that is the same as $\int _{1}^{x}\int _{0}^{\infty }e^{-t}t^{x-1}\operatorname {d} \!t\operatorname {d} \!x=\iint _{R}e^{-t}t^{x-1}\operatorname {d} \!t\operatorname {d} \!x$ (i.e., can we use properties of double integrals, like changing the order of integration, to help us?; R is an appropriate region under the graph of $F(x,t)=e^{-t}t^{x-1}$ on the xt plane)? The answer I'm looking for need not be expressible in terms of elementary functions, and would almost certainly be not, but I'd like something other than an infinite series.--Jasper Deng (talk) 04:47, 24 May 2013 (UTC)[reply]

I'd like something other than an infinite series — Is that so ? :-) 79.113.225.20 (talk) 12:01, 24 May 2013 (UTC)[reply]

Not helpful. I'm looking for an integral expression or some other form that is more elegant than a series.--Jasper Deng (talk) 17:30, 24 May 2013 (UTC)[reply]

Gamma(z) has a simple representation in terms of its poles in the complex plane, you could just integrate that term by term. Count Iblis (talk) 19:20, 24 May 2013 (UTC)[reply]

...which, however, would not lead to something other than an infinite series, at least to the degree I understand it (note - I'm not proficient in complex analysis).--Jasper Deng (talk) 01:35, 25 May 2013 (UTC)[reply]

which, however, would not lead to something other than an infinite series — Told you! :-) But would the following be more to your liking ?

\int {x!}\ dx=\int \int _{0}^{\infty }{t^{x} \over e^{t}}\ dt\ dx=\int _{0}^{\infty }{t^{x} \over {e^{t}\ln t}}\ dt

Also, here are a few alternative expressions for the factorial/Gamma function:

x!=\Gamma (x+1)=\int _{0}^{\infty }{t^{x} \over e^{t}}\ dt=\int _{0}^{\infty }{e^{-{\sqrt[{x}]{t}}}}\ dt=\int _{0}^{1}{|\ln t|^{x}}\ dt=\int _{0}^{1}{(-\ln t)^{x}}\ dt=\int _{0}^{1}{\ln ^{x}{\tfrac {1}{t}}}\ dt=\int _{1}^{\infty }{\ln ^{x}t \over t^{2}}\ dt

Which, when integrated relative to x under the integral sign, yield the following results:

\int {x!}\ dx=\int _{0}^{\infty }{t^{x} \over {e^{t}\ln t}}\ dt=\int _{0}^{1}{|\ln t|^{x} \over \ln |\ln t|}\ dt=\int _{0}^{1}{(-\ln t)^{x} \over \ln(-\ln t)}\ dt=\int _{0}^{1}{\ln ^{x}{\tfrac {1}{t}} \over \ln(\ln {\tfrac {1}{t}})}\ dt=\int _{1}^{\infty }{\ln ^{x}t \over t^{2}\ln(\ln t)}\ dt

Also, you said you disliked infinite sums, but you said nothing about infinite products :-), so:

x!=\Gamma (x+1)=\prod _{n=1}^{\infty }{\frac {\left(1+{\frac {1}{n}}\right)^{x}}{1+{\frac {x}{n}}}}\ =\ e^{-\gamma x}\ \prod _{n=1}^{\infty }{e^{\frac {x}{n}} \over 1+{\frac {x}{n}}}

Hope this helps! :-) — 79.113.226.14 (talk) 07:33, 25 May 2013 (UTC)[reply]

Yes, those integral and infinite product formulations do answer my question.--Jasper Deng (talk) 17:34, 25 May 2013 (UTC)[reply]

Ruler and compass construction edit

I'm interested in how you construct the line parallel to a given line l through a given point A, using only ruler and compass. I'm aware of the following methods, which I'll state in the barest outline:

1. Construct the perpendicular m to l through A, and then the perpendicular to m through A. (Until I became interested in this question recently, this would have been my natural answer, probably because I've been familiar with the method for constructing a perpendicular since childhood.)

2. Construct a parallelogram with one side on l and one vertex at A, using the fact that opposite sides of a parallelogram have equal lengths. I've only seen rhombuses used for this so far.

3. Draw any line m through A (not parallel to l!) and reproduce, at A, the angle formed by m with l.

4. Construct an isosceles triangle with base on l for which A will be the midpoint of one side. Then the line connecting the midpoints not on l is the desired line.

Are there other well-known or elegant methods? Is there a particular method that is considered "standard" in some way (whether for historical reasons or objective ones)? 96.46.198.58 (talk) 05:15, 24 May 2013 (UTC)[reply]

One method that comes to the top of my head is to construct two circles on l of equal radii and then draw a line tangent to both circles using your straightedge.--Jasper Deng (talk) 05:22, 24 May 2013 (UTC)[reply]

How do you get the line to go through A? Also, presumably, pushing the straightedge up against the two circles isn't a permitted step, so this would need to be done indirectly, by constructing perpendiculars. 96.46.198.58 (talk) 05:33, 24 May 2013 (UTC)[reply]

5. (Invented right now.) Find on l a base of an isosceles triangle with apex A (just an arc with center A), then shift the base along l by arbitrary distance and reconstruct a sibling triangle with apex B – then AB is the line you seek. --CiaPan (talk) 06:45, 24 May 2013 (UTC)[reply]

Update: Shifting is not allowed, the compass collapses when lifted from the plane. So let's call endpoints of the first triangle's base C and D, then the second base is found as CE with circle centered at C with radius CD. --CiaPan (talk) 06:59, 24 May 2013 (UTC)[reply]

Thanks. I like this construction. I don't mind copying distances, as I've always understood this to be a permitted operation. I know that mathematically, deciding what is permitted is arbitrary, but my question is partly mathematical and partly cultural. 96.46.198.58 (talk) 13:33, 25 May 2013 (UTC)[reply]

It turned out that copying lengths is possible (even with the compass alone), although not allowed! See Compass equivalence theorem – one may not shift a line segment, however one can construct a new point (F in first, E in second method), which is a given distance (BC) apart from some given point A. --CiaPan (talk) 05:22, 27 May 2013 (UTC)[reply]

Well I'd do it by drawing three circles, one through A intersecting the line twice, then another of the same length from one of those intersection points, and then another from A of the distance between the two intersection points. Then I just draw a line through A and the last intersection point I get. I think that corresponds most closely to method 2. Dmcq (talk) 08:38, 24 May 2013 (UTC)[reply]

This is a version of method 2 that uses a parallelogram which usually won't be a rhombus. I noticed this exact construction as an alternative to the rhombus method. 96.46.198.58 (talk) 13:33, 25 May 2013 (UTC)[reply]

Interestingly being able to copy a length like I do for that last circle is not needed for compass and straightedge constructions, one can always do that some other way just using circles where the centre and a redius point are already present Dmcq (talk) 10:41, 24 May 2013 (UTC)[reply]

A better solution which doesn't require copying lengths: draw a circle through A cutting the line at B and C. With centre B draw a line with radius BA, with centre C draw a circle radius CA (these circles are all equal radius). The last two circles meet at D on the opposite side of the line. Draw a line through D and C meeting the circle centre C at E. Then the line AE is parallel to the original line. Dmcq (talk) 10:51, 24 May 2013 (UTC)[reply]

Thanks. I like this too. 96.46.198.58 (talk) 13:33, 25 May 2013 (UTC)[reply]

I'm still wondering if there is a particular way of doing this that has traditionally been taught in schools as "the" way you do it. That was part of my question too. If anybody can point out a reference that might have answers, that would be much appreciated. 96.46.198.58 (talk) 13:33, 25 May 2013 (UTC)[reply]

6) Choose (almost) any point B at line l and draw a semicircle centered at B passing through A. Its endpoints at line l are C and D. Pick the distance CA and draw an arc centered at D with that radius. The point E where it meets the first arc defines together with A the line parallel to l (and ACDE, or ADCE, is an isosceles trapezoid).

7) Modification of 6) if you avoid picking and transferring lengths: choose point B in l and draw a semicircle, as in 6); that is arc 1; choose its endpoint C and draw a semicircle centered at C from endpoint B – that is arc 2; mark its endpoint other than B with label D and draw a circle with center D passing C – arc 3; finally draw an arc with center C passing through A – one of points where it meets arc 3 defines your line.

CiaPan (talk) 10:19, 27 May 2013 (UTC)[reply]

And here's yet another. From two different points on the line draw circles through point A. Through the other intersection point of the two circles draw a straight line through the centre of one of the circle to intersect thecircle on the opposite side. From where that line intersects the opposite side draw a line through A. That gives you the parallel line. Dmcq (talk) 22:58, 27 May 2013 (UTC)[reply]

To try and answer your question about a standard method: I had a quick look and as far as I an see your method 3 seems to be by far the most common. However they move the compass to copy a length so it doesn't comply with the strict requirements of straightedge and compass. Dmcq (talk) 19:03, 28 May 2013 (UTC)[reply]

As I mentioned above, there is a solution: Compass equivalence theorem (however it makes the construction much more complicated). --CiaPan (talk) 05:25, 29 May 2013 (UTC)[reply]

Why only 2 tools? edit

I've always wondered why they always say "only ruler and compass". Seems to me there's a third tool required, a writing implement. Or, if you're doing it on sand or dirt, a stick or your finger. I suppose you could use the end of the ruler in that case too, but I sense that that use of it was not anticipated. Any ideas? -- Jack of Oz ^[Talk] 06:02, 24 May 2013 (UTC)[reply]

A ruler is just any tool to draw a line thgouh two given points. See Compass-and-straightedge construction for detailed description of the two tools' properties (they are essentially implementation of three Euclid's postulates). --CiaPan (talk) 06:52, 24 May 2013 (UTC)[reply]

I don't get that. How can a ruler or a straightedge "draw" a line, unless as I said it was all done on sand or some such material? The pencil in the compass seems to be taken for granted. Is that the same instrument used to actually draw lines in these constructions? -- Jack of Oz ^[Talk] 07:26, 24 May 2013 (UTC)[reply]

You're taking it too literally. It's like saying that to evaluate polynomials only requires addition and multiplication; it's abstract, there's no need to mention the pencil and paper to record results.Phoenixia1177 (talk) 08:05, 24 May 2013 (UTC)[reply]

In the polynomials case, addition and subtraction are the only mathematical tools you need. But the straightedge and compass constructions use real, physical objects to make them work. If all you had was a compass and a ruler, and no pencil, you could not make the constructions. They rely on lines and sections of circles actually being drawn, and it's important to identify exactly the intersections of those lines. You can't do any of that without some writing implement. The pencil may be taken for granted in the polynomial case, but I can't see that that is true in this case. It's as fundamentally important as the straightedge and the compass, but it gets no billing. I'm not being wilfully obtuse here. -- Jack of Oz ^[Talk] 08:54, 24 May 2013 (UTC)[reply]

Again, you're taking this to literally. The pencil adds nothing to the mathematics; the point isn't a physical compass and straight-edge, but theoretical ones; the same way that Turing Machines don't refer to the mechanics of the device to feed the tape. Mentioning the pencil, mathematically, is as important as mentioning the material the compass is made from. Further evidence that it is all abstract is that there is no bound on the sizes of the compass or straight-edge. Essentially, the tools are just a reference to the actual abstract operations.Phoenixia1177 (talk) 09:28, 24 May 2013 (UTC)[reply]

Construction of a square.

Well, if we're actually talking about a theoretical straightedge and a theoretical compass (concepts that had never entered my consciousness before, despite 3 years of university mathematics), why can we not have theoretical lines drawn by a theoretical pencil? -- Jack of Oz ^[Talk] 10:15, 24 May 2013 (UTC)[reply]

To what end? What does that add? Why don't we discuss what material our turning machines are made from? Also, I don't mean this rudely, but how many years of university mathematics you've had doesn't matter; and three isn't even enough to make you an expert at mathematics, so it's especially irrelevant. As for "never entering your consciousness before" that they were theoretical, a real world compass and straight edge would have defects that would make the whole affair pointless.Phoenixia1177 (talk) 10:20, 24 May 2013 (UTC)[reply]

I never claimed to be an expert; I was just making the point that I was not aware that this theory was never meant to be applied in practice, which makes a bit of a mockery of the moving image in the right, borrowed from our article. If that's not a set of instructions on how to actually, physically draw a square on an actual, physical piece of paper, what is it? It shows three devices: a straightedge, a compass and a pencil. I am no closer to understanding why the theory only mentions the first two of these. -- Jack of Oz ^[Talk] 11:13, 24 May 2013 (UTC)[reply]

It applies to actually carrying out the construction in the same way triangles exist in the real world and the same way computers are "equivalent" to turing machines. Speaking of our article, the first few sentences mention the tools being idealized. All that aside, though, how could you prove theorems about nontheoretical objects? And all that aside, what does the pencil add? And why not a pen? Why not a laser beam that etches in wood? It's not mentioned because it adds nothing to the mathematics of the situation, it is a trivial detail. We don't care about the pencil for the same reason we don't care about the turing machines tape feeder.Phoenixia1177 (talk) 11:21, 24 May 2013 (UTC)[reply]

In my opening question I did call it a "writing implement". Not sure what you mean by triangles existing in the real world. They probably occur in nature in various ways, but if we want to create a new one, we have to draw it with a writing implement or construct it using physical materials. -- Jack of Oz ^[Talk] 11:28, 24 May 2013 (UTC)[reply]

Triangles don't occur at all, you can't draw one. You can't get a hold of the infinite ruler mentioned either. And I've repeatedly answered your question: the theoretical writing implement is not mentioned because it has no mathematical value.Phoenixia1177 (talk) 11:36, 24 May 2013 (UTC)[reply]

Thanks for your efforts. Maybe it needs someone else to explain it to me in a way I can understand. -- Jack of Oz ^[Talk] 11:46, 24 May 2013 (UTC)[reply]

I agree.Phoenixia1177 (talk) 11:48, 24 May 2013 (UTC)[reply]

If I can weigh in, here's why I think Pheonixia is right. Basically, redundancy is antithetical to mathematical pursuits. Specifying a writing implement is a redundancy -- it's obvious that for a real-world attempt at the construction, such a tool must be on hand. The same is not true for the two other tools, because their presence actually impacts what you can construct. —Anonymous Dissident^Talk 11:51, 24 May 2013 (UTC)[reply]

A writing instrument is part of a ruler or compass. In the real world one cannot use a ruler or compass without a writing instrument. Obviously a writing instrument need not be a pencil or pen; writing in the sand requires a "writing instrument". There are many ways of "writing" but one can't eliminate all of them and still say that one has a ruler or compass. Properly defined, a real world ruler or compass includes some form of "writing instrument". Bus stop (talk) 11:53, 24 May 2013 (UTC)[reply]

OK, I think I'm starting to get this. The pen and the ruler go hand in hand and are considered to be parts of the same device, for the purposes of this theory (even though they're clearly separate objects in real life). This raises two questions: (a) If only one of these items is going to be mentioned, why is it the ruler and not the pencil? (b) Why do we need the ruler at all? Answer: Because humans are really bad at drawing perfectly straight lines freehand. We need the ruler to make sure the line drawn by the pencil is straight. If one could train oneself to draw a perfectly straight line freehand, one could dispense with the ruler. But it would never be possible to dispense with the pencil. That leaves the pencil at least as important as the ruler in this scenario. Back to Square 1. -- Jack of Oz ^[Talk] 12:12, 24 May 2013 (UTC)[reply]

Eraser is also handy. Gzuckier (talk) 17:07, 24 May 2013 (UTC)[reply]

It is not inconceivable that a human could draw a straight line with less deviation from perfection than could be accomplished using a straightedge. In that case the proper name of that individual might come to symbolize straight lines. If another human could supplant the role of a compass in drawing circles, one could say "draw a line parallel to line A-B using only a Bob and a Joe," assuming these were the names of the individuals possessing those prodigious skills. Under these arrangements I think it is logical that Bob and Joe would possess pencils. Bus stop (talk) 15:07, 24 May 2013 (UTC)[reply]

An alternative point: the pencil's job is to create. But mathematically, we're not creating lines and circles; in fact, I'm not even sure what that would mean. Instead, we're identifying existing lines and circles. We put down the ruler between two points, see the line which runs along the ruler, and say "yes, that's the line I'm looking for".--149.148.255.22 (talk) 13:22, 24 May 2013 (UTC)[reply]

It's consistent with Plato's philosophy that the circle and line is already there. The function of the compass and straightedge is just to indicate lines in the sand to remind us where those actual circles and lines are. Sławomir Biały (talk) 14:21, 24 May 2013 (UTC)[reply]

Euclid talked about a line having length but not breadth. However that doesn't mean a line goes from one point to another and so has a particular length - that is a line segment. A line is infinite (or potentially infinite as the Greeks might have said in that it can be extended arbitrarily far). So in the case of 'a Bob and a Joe' they would take an awfully long time drawing an actual line using a pencil! Dmcq (talk) 15:44, 24 May 2013 (UTC)[reply]

If these circles and lines are already there, why did Phoenixia say "Triangles don't occur at all, you can't draw one"? -- Jack of Oz ^[Talk] 20:24, 24 May 2013 (UTC)[reply]

Because I was talking about finding triangles in the "real world", this is not the same sense as saying they exist as a Platonist. In other words, if you walk outside of your home and look around, you will not see any triangles; you will see approximations of triangles. What I was trying to say is that actually drawing stuff with a ruler and straight-edge relates to the theory in the same way, it is an approximation not the mathematical point.Phoenixia1177 (talk) 21:10, 24 May 2013 (UTC)[reply]

Well this was a shortcoming of Greek philosophy. For the Greeks everything that existed in Plato's sense was constructable. Most modern ontologies of mathematics reject this edict. Sławomir Biały (talk) 21:52, 24 May 2013 (UTC)[reply]

I didn't know that; I thought Plato's forms also covered things like truth and beauty, things that are not "real" in the "grab a hold of it" sense. I've read some on the modern variants, not much on the older; anything you can recommend that gets into it?Phoenixia1177 (talk) 10:04, 25 May 2013 (UTC)[reply]

Interior of connected components edit

Consider a topological space X. Is it always the case that for p in X, the connected component of p has a connected interior? It's clear for connected spaces. I know it is the case for locally connected spaces, since in such cases the connected component is open. It's also true for totally disconnected spaces, like the rationals with the subspace topology, because the connected components are singletons with empty interior. So a candidate counterexample would need to be neither connected nor locally connected, but not totally disconnected. Do such spaces even exist? —Anonymous Dissident^Talk 10:40, 24 May 2013 (UTC)[reply]

Okay, an example is a comb space plus one point outside. That doesn't contradict the original claim though... —Anonymous Dissident^Talk 11:05, 24 May 2013 (UTC)[reply]

No. Consider the space

\{a,b,c\}\times \mathbb {Q}

, with the following basic open sets:

\{(a,q)\}

for any

q\in \mathbb {Q}

;

\{(b,q)\}

for any

q\in \mathbb {Q}

; and

\{a,b,c\}\times I

for any open interval

I\subseteq \mathbb {Q}

. Because the rationals are totally disconnected, the connected component of any point is its triple

\{(a,q),(b,q),(c,q)\}

, for the appropriate

q

. But the interior of a triple is

\{(a,q),(b,q)\}

, which is discrete.--149.148.255.22 (talk) 12:17, 24 May 2013 (UTC)[reply]

If you want, this space can be made Hausdorff by replacing

a

and

b

with intervals. Take the space

(-1,1)\times \mathbb {Q}

, with the following basic open sets:

I\times \{q\}

for any

q\in \mathbb {Q}

and any open interval

I\subset (-1,1)

which does not contain 0; and

I\times J

for any open intervals

I\subseteq (-1,1),J\subseteq \mathbb {Q}

. Again, the connected component of a point is

(-1,1)\times \{q\}

, but the middle of the interval is in the boundary.--149.148.255.22 (talk) 12:27, 24 May 2013 (UTC)[reply]

I haven't investigated the Hausdorff example, but I'm not sure I follow with the first one. For instance, is it not the case that the set

\{(a,q),(b,q),(c,q),(c,p)\}

is connected for any

p\neq q

, so that the connected component you gave is not right? —Anonymous Dissident^Talk 07:25, 27 May 2013 (UTC)[reply]

No, for the same reason the rationals are totally disconnected. Fix any irrational

\beta

with

q<\beta <p

(without loss of generality,

q<p

). Then

\{a,b,c\}\times \mathbb {Q} _{<\beta }

and

\{a,b,c\}\times \mathbb {Q} _{>\beta }

are disjoint open sets which cover the space, and they separate

(c,q)

from

(c,p)

.--149.148.254.116 (talk) 15:25, 27 May 2013 (UTC)[reply]

Ah, I forgot how the topology subspace works (or, the definition of connectedness for subspaces; either way). Silly me. Thanks for your help; this is a really compelling counterexample. Is it a standard one? If not, could you elaborate on how you concocted it? Can we generalise the result to talk about exactly which spaces have the property that there are points having a connected component with a disconnected interior? —Anonymous Dissident^Talk 00:04, 28 May 2013 (UTC)[reply]

It wouldn't surprise me if Counterexamples in Topology had an example along these lines, but I came up with it on my own. I started by finding a set which was connected, but its interior was disconnected; the example I came up with was formed by two crescents with their tips touching. Then I realized I could simplify that significantly. The idea of attaching copies of spaces at a critical point so that any neighborhood of the critical point must bleed into other copies comes from the wedge of circles. I just needed to modify it to keep separate copies in separate components.

I doubt there will be any simpler classification of such spaces.--149.148.255.22 (talk) 09:26, 28 May 2013 (UTC)[reply]

Combining p-values edit

How might one go about combining p-values (in a null hypothesis significance testing sense) when there are multiple different experiments of substantially different character? For example, how would you go about determining if a casino is fair - in an overall sense - if there's multiple games (say coin flips, dice rolls and roulette wheel spins)? Given the appropriate data, it's straightforward to get p-values for each of the games, but how would you combine them into a single significance test for the casino as a whole? (In my particular case, it isn't coin flips/dice rolls, but rather before/after comparisons of several different data sets. I can get p-values for each set individually through permutation testing, but each set is on completely different scales, so aggregating them into a single set doesn't make much sense.) Is there an accepted method to do this? -- 205.175.124.30 (talk) 22:06, 24 May 2013 (UTC)[reply]

This is an instance of the problem of multiple comparisons. The simplest approach is to do a Bonferroni correction, but as our article explains, there might be more sophisticated approaches you could use. Looie496 (talk) 23:23, 24 May 2013 (UTC)[reply]

My understanding was that the Bonferroni correction and similar multiple comparison compensations were for multiple different tests on the same data. My situation is the inverse of that - I'm looking at a single hypothesis test on several different datasets simultaneously. That is, I'm not interested in testing if any single subset when taken independently rejects the null hypothesis, but interested in a test that considers all the subsets taken as a whole. I'm not sure if/how multiple comparison corrections would be applied. -- 205.175.124.30 (talk) 23:59, 24 May 2013 (UTC)[reply]

No, the Bonferroni correction is so simple-minded that it works even for unrelated tests. (Actually the less related they are the better -- it's often too conservative when applied to nonindependent tests.) The Bonferroni rule is that if you do N tests, of whatever sort, and wish to get an overall significance level of p, then you should require that at least one of the tests be significant at level p/N. So if, for example, you have 10 games and want to reject the hypothesis that the casino is fair at a significance level of p < 0.01, then you must be able to reject the hypothesis that one of the games is fair at a level p < 0.001. Looie496 (talk) 01:24, 25 May 2013 (UTC)[reply]

Okay, I think I have a better understanding now. - Though what about a case where you're slightly more forgiving of the casino? That is, instead of rejecting the null hypothesis that the casino is fair as a whole if any one of its games is unfair (which is what I understand the Bonferroni is effectively doing), you instead only reject the overall null hypothesis on the casino if all of the individual games are unfair. (Or perhaps some intermediate level, like 3/5 games are unfair.) Would I be correct in assuming that if you truly had independence in the games, the way to handle it would be to apply the Bonferroni on each game individually, and then see if the accept/rejects exceed your ratio? - It doesn't seem quite right to me for some reason, perhaps because I'm conflating it with the scenario where things *aren't* independent. Specifically the situation where you can impose the external knowledge that if the casino is fair, it should be fair for all games, and if it's unfair, then it will be unfair for all games (so using loaded dice but fair coins and fair roulette wheels isn't a valid alternate hypothesis.). Is there a way to combine p-values under this all-or-nothing condition? -- 205.175.124.30 (talk) 21:29, 25 May 2013 (UTC)[reply]

A null hypothesis needs to be detailed enough to generate probability distributions for the statistics you are testing. A statement such as "all the games are fair" is a proper null hypothesis, because, presuming you understand how every game works, it allows you, at least in principle, to generate a probability distribution for any test statistic you might come up with. However, statements such as "all the games are unfair" or "three quarters of the games are unfair" are not proper null hypotheses, unless you specify the exact type of unfairness. There may be some way of handling them, but you would have to do unusual things (e.g., Bayesian statistics). Anyway, these complications make the problem too weird for me to know how to cope with. Looie496 (talk) 03:01, 26 May 2013 (UTC)[reply]

You should use a modified Bonferroni correction such as the Hochberg correction: it is less conservative than the Bonferroni correction, while still controlling the experiment-wise error at a given alpha. For your second question--rejecting the overall null hypothesis if you reject the null hypotheses for a given proportion of casino games--my first thought is to model the overall experiment as a series of binomial trials, and then test the null hypothesis that a given number of trials would have their null hypotheses rejected. The problem here is that you would need to specify the proportion prior to the test, and justify your choice; I think it will probably be more convincing to just argue for your overall conclusion rather than try to justify it statistically in this manner.OldTimeNESter (talk) 08:04, 27 May 2013 (UTC)[reply]

martingale versus random walk edit

Let's say you have $950 in play money in a casino game, doesn't matter what you started with. The rule is if you get to $1000 in play money you get to keep a prize (car), beyond that limit - which you're just trying to hit - it doesn't matter what play money you have. However if you lose all your play money you have to stop playing.

What is the optimal way of increasing the odds that you make it to $1000 at least once - should you just take a random walk from $950 by betting the minimum? Or should you try martingaling, for example, betting $50, then if you lose (down to $900) betting $100, then if oyu lose (down to $800) betting $200, then if you lose (down to $600) betting $400, then if you lose (down to $200) which is not enough for another martingale, just keep betting it all?

Or is there some other optimal strategy? or it doesn't matter?

Assume you can bet any amount, but the house has some finite advantage on each bet. 178.48.114.143 (talk) 23:16, 24 May 2013 (UTC)[reply]

If the house has an advantage, in most cases your best strategy will be the one that gets you to $1000 with the minimum number of bets. I think the only situation where that doesn't hold is if the house's advantage increases more than linearly as a function of the size of the bet. Looie496 (talk) 23:30, 24 May 2013 (UTC)[reply]

What is that supposed to be? The minimum number of possible bets, or average bets, or what? Can you be precise? Assume a fixed percentage house advantage, you can bet any amount you want (to any precision) up to the amount you have in money. --178.48.114.143 (talk) 23:54, 24 May 2013 (UTC)[reply]

Well, that depends on details that you haven't specified. Generally speaking, the more bets you make, the more likely it is that your return will be close to the average return. Therefore, as the number of bets increases, the probability of being ahead after than number of bets drops exponentially (actually faster than exponentially). Thus, the basic principle is that you either have to get there quickly or you won't get there at all. Looie496 (talk) 01:29, 25 May 2013 (UTC)[reply]

Wikipedia:Reference desk/Archives/Mathematics/2013 May 24

Contents

May 24 edit

Integral of different type edit

Antiderivative of gamma function edit

Ruler and compass construction edit

Why only 2 tools? edit

Interior of connected components edit

Combining p-values edit

martingale versus random walk edit